IMO 1964

Key idea: $2^3=8\equiv1\pmod7$, so $2^n$ modulo $7$ only depends on $n$ modulo $3$.

Write $n=3q+r$ with $r\in\{0,1,2\}$. Then $2^n=8^q\cdot2^r\equiv2^r\pmod7$, so $2^n$ leaves remainder $1$, $2$ or $4$ for $r=0,1,2$.

(a) $7\mid2^n-1$ exactly when the remainder is $1$, that is when $3\mid n$.

(b) $2^n+1$ leaves remainder $2$, $3$ or $5$, never $0$. So $7$ never divides $2^n+1$.

Key idea: after expanding, this is Schur’s inequality, which holds for all non-negative numbers, not just triangle sides.

Expanding, the left side is $a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)$. So the claim is $$a^3+b^3+c^3+3abc\ \ge\ a^2(b+c)+b^2(c+a)+c^2(a+b),$$ which can be rewritten as $$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\ge0 .$$ (Expand each product to check.) By symmetry, assume $a\ge b\ge c\ge0$. The last term is $c(a-c)(b-c)\ge0$. The first two combine to $$(a-b)\bigl[a(a-c)-b(b-c)\bigr]\ge0,$$ because $a-b\ge0$ and $a(a-c)\ge b(b-c)$ (each factor on the left is at least the matching factor on the right, and all are non-negative). So the sum is non-negative.

For a triangle, equality holds only when $a=b=c$: in the ordered case, equality needs $c(a-c)(b-c)=0$ and $(a-b)[\dots]=0$, and for positive sides this forces $a=b=c$.

Key idea: each corner triangle is a scaled copy of $ABC$, with factor $\frac{s-a}{s}$. So its incircle has radius $\frac{s-a}{s}\rho$.

Let $s=\frac{a+b+c}2$, let $\rho$ be the inradius and $T$ the area. The tangent parallel to $BC$ (other than $BC$ itself) is at distance $2\rho$ from $BC$, on the side of $A$. It cuts off a triangle at $A$ that is similar to $ABC$, with ratio of heights $$k_a=\frac{h_a-2\rho}{h_a}=1-\frac{2\rho}{h_a}=1-\frac{2T/s}{2T/a}=\frac{s-a}{s}.$$ Its incircle has radius $k_a\rho$, and similarly at $B$ and $C$. The total area is $$\pi\rho^2\Bigl(1+\frac{(s-a)^2+(s-b)^2+(s-c)^2}{s^2}\Bigr).$$ Expanding, $(s-a)^2+(s-b)^2+(s-c)^2=3s^2-2s(a+b+c)+a^2+b^2+c^2=a^2+b^2+c^2-s^2$, because $a+b+c=2s$. So the bracket is $\frac{a^2+b^2+c^2}{s^2}$. Heron’s formula gives $\rho^2=\frac{T^2}{s^2}=\frac{(s-a)(s-b)(s-c)}{s}$. So the total is $$\frac{\pi(s-a)(s-b)(s-c)(a^2+b^2+c^2)}{s^3}.$$ With $x=b+c-a$, $y=c+a-b$, $z=a+b-c$ (so $s-a=\frac x2$, and so on) this is $$\frac{\pi\,xyz\,(a^2+b^2+c^2)}{(a+b+c)^3}.$$ For an equilateral triangle of side $1$ this is $\frac{\pi}{9}$: the big incircle has area $\frac{\pi}{12}$ and each small one $\frac{\pi}{108}$.

Key idea: pigeonhole twice, the same idea that shows any party of six has three mutual friends or three mutual strangers.

Think of people as points and color the line between two people by their topic: three colors.

Six people, two colors. Pick one person $P$ among six. Of the $5$ lines from $P$, at least $3$ have the same color, say red, going to $Q,R,S$. If any line among $Q,R,S$ is red, it forms a red triangle with $P$. If none is, $QRS$ is a triangle in the other color.

Seventeen people, three colors. Pick one person $P$. The $16$ lines from $P$ use three colors, so at least $6$ share a color, say red (since $3\cdot5=15<16$). Call those six people the group.

• If two people in the group write to each other in red, they and $P$ form a red triangle.
• Otherwise the six people in the group use only the other two colors among themselves, and by the first step they contain a one-color triangle.

Either way, some three people all share one topic in their letters.

Key idea: start from all pairs of perpendiculars and subtract the coincidences that geometry forces. Then exhibit a configuration where nothing else coincides.

Each point is joined to the other four by $\binom42=6$ lines not through it, so there are $5\cdot6=30$ perpendiculars and $\binom{30}2=435$ pairs of them.

Forced coincidences.

1. Parallel pairs. The perpendiculars to the same joining line are parallel. Each of the $10$ joining lines receives perpendiculars from the $3$ points not on it, giving $3$ parallel pairs. That is $30$ pairs with no intersection.
2. Pairs through a given point. The $6$ perpendiculars from the same point meet only there: $15$ pairs give $1$ point. Over five points, $75$ pairs give only $5$ points.
3. Altitudes. For each of the $\binom53=10$ triangles, its three altitudes are perpendiculars (from a vertex to the opposite side’s line) and meet at the orthocenter. Their $3$ pairs give $1$ point. These pairs join perpendiculars from different points to non-parallel lines, so they are not counted in 1 or 2. That is $30$ pairs giving $10$ points.

So the number of distinct intersection points is at most $$(435-30-75-30)+5+10=300+15=315 .$$

The bound is reached. We need one admissible configuration with no further coincidence. The points $(0,0)$, $(0,2)$, $(2,5)$, $(6,8)$, $(7,6)$ qualify: none of the $10$ joining lines are parallel, perpendicular or equal. Computing all intersection points of the $30$ perpendiculars in exact rational arithmetic gives exactly $315$ distinct points. (A configuration in general position, for instance one chosen at random, does the same.)

So the maximum is $315$.

Key idea: with barycentric coordinates for $D_0$, each new point sits above a vertex at height $h/\alpha$, $h/\beta$ or $h/\gamma$. The determinant then adds up three areas weighted by $\alpha,\beta,\gamma$ and cancels these weights exactly.

Let $D_0=\alpha A+\beta B+\gamma C$ with $\alpha,\beta,\gamma>0$ and $\alpha+\beta+\gamma=1$. (The centroid has all three equal to $\frac13$, but we keep them general.) A shear that fixes the plane $ABC$ and moves $D$ parallel to that plane preserves volumes, parallel lines and planes. So we may assume $\vec e=\overrightarrow{D_0D}$ is perpendicular to the plane $ABC$, of length $h$.

The new points. A point $A+t\vec e$ lies in plane $BCD$ when $A+t\vec e=xB+yC+zD$ with $x+y+z=1$. Since $D=\alpha A+\beta B+\gamma C+\vec e$, comparing the parts along $\vec e$ and the coefficients of $A,B,C$ gives $t=z$ and $1=z\alpha$. So $t=\frac1\alpha$, and $$A_1=A+\tfrac1\alpha\vec e,\qquad B_1=B+\tfrac1\beta\vec e,\qquad C_1=C+\tfrac1\gamma\vec e .$$ The volume. Put $D_0$ at the origin and let $\mathbf a,\mathbf b,\mathbf c$ be the positions of $A,B,C$ in the plane. Let $S$ be the area of $ABC$. Six times the volume of $A_1B_1C_1D_0$ is the absolute value of the determinant of the three vectors $\overrightarrow{D_0A_1},\overrightarrow{D_0B_1},\overrightarrow{D_0C_1}$. Their parts in the plane are $\mathbf a,\mathbf b,\mathbf c$ and their heights are $h/\alpha,h/\beta,h/\gamma$. Expanding the determinant along the heights gives $$\frac{h\,(\mathbf b\times\mathbf c)}{\alpha}+\frac{h\,(\mathbf c\times\mathbf a)}{\beta}+\frac{h\,(\mathbf a\times\mathbf b)}{\gamma},$$ with $\times$ the planar cross product. Now $\mathbf b\times\mathbf c$ is twice the signed area of $D_0BC$, which is $2\alpha S$ by the meaning of barycentric coordinates. Likewise $\mathbf c\times\mathbf a=2\beta S$ and $\mathbf a\times\mathbf b=2\gamma S$. So the determinant is $6hS$, and $$\operatorname{vol}(A_1B_1C_1D_0)=hS=3\cdot\tfrac13hS=3\operatorname{vol}(ABCD).$$ The weights cancelled for every interior point $D_0$, not just the centroid. So the answer to the last question is yes.